Sketch an angle $\theta$ in standard position such that $\theta$ has the least possible positive measure, and the point $(-3, \sqrt{2})$ is on the terminal side of $\theta$. Then find the values of the six trigonometric fun angle. Rationalize denominators.

Step 1 ：Sketch an angle \(\theta\) in standard position such that \(\theta\) has the least possible positive measure, and the point \((-3, \sqrt{2})\) is on the terminal side of \(\theta\).

Step 2 ：The point \((-3, \sqrt{2})\) lies in the second quadrant. The angle \(\theta\) in standard position with the least positive measure that has this point on its terminal side is the angle formed by the negative x-axis and the line segment connecting the origin to the point \((-3, \sqrt{2})\).

Step 3 ：To find the values of the six trigonometric functions, we first need to find the length of the hypotenuse of the right triangle formed by the x-axis, the line segment connecting the origin to the point \((-3, \sqrt{2})\), and the line segment from the point \((-3, \sqrt{2})\) perpendicular to the x-axis. This is given by the distance formula, which in this case simplifies to the Pythagorean theorem.

Step 4 ：The six trigonometric functions are then given by the ratios of the lengths of these sides.

Step 5 ：\(\sin(\theta) = \frac{\sqrt{2}}{\sqrt{11}}\), \(\cos(\theta) = -\frac{3}{\sqrt{11}}\), \(\tan(\theta) = -\frac{\sqrt{2}}{3}\), \(\csc(\theta) = \frac{\sqrt{11}}{\sqrt{2}}\), \(\sec(\theta) = -\frac{\sqrt{11}}{3}\), \(\cot(\theta) = -\frac{3}{\sqrt{2}}\)

Step 6 ：\(\boxed{\sin(\theta) = \frac{\sqrt{2}}{\sqrt{11}}, \cos(\theta) = -\frac{3}{\sqrt{11}}, \tan(\theta) = -\frac{\sqrt{2}}{3}, \csc(\theta) = \frac{\sqrt{11}}{\sqrt{2}}, \sec(\theta) = -\frac{\sqrt{11}}{3}, \cot(\theta) = -\frac{3}{\sqrt{2}}}\)